Graphing Parabolas Given in General Form

Question 1 of 4

Sketch the graph of

$x^2+2x-4y-11=0$

1.
2.
3.
4.

Incorrect

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Standard Form (Concave Up)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,$ $a$

$+$ $k$

$)$

Standard Form (Concave Down)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,-$ $a$

$+$ $k$

$)$

First, transform the given equation into vertex form.

$x^2+2x-4y-11$	$=$	$0$
$x^2+2x$	$=$	$4y+11$	Keep only $x$ terms on the left side
$x^2+2x$ $+1$	$=$	$4y+11$ $+1$	Add $1$ to both sides to complete the square
$(x+1)^2$	$=$	$4y+12$	Contract left side
$(x+1)^2$	$=$	$4(y+3)$	Factor right side

Now, determine which standard form applies to the given equation.

$(x+1)^2$

$=$

$4(y+3)$

Since the coefficient of

$(y+3)$ is positive, use

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

This also means that the parabola will be concave up

Find the focal length ( $a$ ) and vertex by comparing the given equation to the standard form

$(x-\color{#00880a}{h})^2$	$=$	$4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$
$(x-\color{#00880a}{(-1)})^2$	$=$	$4(y-\color{#007ddc}{(-3)})$
$(x+1)^2$	$=$	$4(y+3)$

The Vertex

$($ $h$

$,$ $k$

$)$ can be read from the equation

$($ $-1$

$,$ $-3$

$)$

Focal Length:

$4a$	$=$	$4$
$4a$ $divide4$	$=$	$4$ $divide4$	Divide both sides by $4$
$a$	$=$	$1$

Identify the focus and directrix.

Focus:

$($ $h$

$,$ $a$

$+$ $k$

$)$ becomes

$($ $-1$

$,$ $1$

$+$ $(-3)$

$)$ or

$(-1,-2)$

Directrix:

$y=-$ $a$

$+$ $k$ becomes

$y=-$ $1$

$+$ $(-3)$ or

$y=-4$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave up

Question 2 of 4

Sketch the graph of

$x^2+4x+8y-4=0$

1.
2.
3.
4.

Incorrect

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Standard Form (Concave Up)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,$ $a$

$+$ $k$

$)$

Standard Form (Concave Down)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,-$ $a$

$+$ $k$

$)$

First, transform the given equation into vertex form.

$x^2+4x+8y-4$	$=$	$0$
$x^2+4x$	$=$	$-8y+4$	Keep only $x$ terms on the left side
$x^2+4x$ $+4$	$=$	$-8y+4$ $+4$	Add $4$ to both sides to complete the square
$(x+2)^2$	$=$	$-8y+8$	Contract left side
$(x+2)^2$	$=$	$-8(y-1)$	Factor right side

Now, determine which standard form applies to the given equation.

$(x+2)^2$

$=$

$-8(y-1)$

Since the coefficient of

$(y-2)$ is negative, use

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

This also means that the parabola will be concave down

Find the focal length ( $a$ ) and vertex by comparing the given equation to the standard form

$(x-\color{#00880a}{h})^2$	$=$	$-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$
$(x-\color{#00880a}{(-2)})^2$	$=$	$-8(y-\color{#007ddc}{1})$
$(x+2)^2$	$=$	$-8(y-1)$

The Vertex

$($ $h$

$,$ $k$

$)$ can be read from the equation

$($ $-2$

$,$ $1$

$)$

Focal Length:

$-4a$	$=$	$-8$
$-4a$ $divide-4$	$=$	$-8$ $divide-4$	Divide both sides by $-4$

$a$	$=$	$2$

Identify the focus and directrix.

Focus:

$($ $h$

$,-$ $a$

$+$ $k$

$)$ becomes

$($ $-2$

$,-$ $2$

$+$ $1$

$)$ or

$(-2,-1)$

Directrix:

$y=$ $a$

$+$ $k$ becomes

$y=$ $2$

$+$ $1$ or

$y=3$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave down

Question 3 of 4

Sketch the graph of

$y^2+2y-12x+13=0$

1.
2.
3.
4.

Incorrect

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Standard Form (Concave Right)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$($ $a$

$+$ $h$

$,$ $k$

$)$

Standard Form (Concave Left)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$(-$ $a$

$+$ $h$

$,$ $k$

$)$

First, transform the given equation into vertex form.

$y^2+2y-12x+13$	$=$	$0$
$y^2+2y$	$=$	$12x-13$	Keep only $y$ terms on the left side
$y^2+2y$ $+1$	$=$	$12x-13$ $+1$	Add $1$ to both sides to complete the square
$(y+1)^2$	$=$	$12x-12$	Contract left side
$(y+1)^2$	$=$	$12(x-1)$	Factor right side

First, determine which standard form applies to the given equation.

$(y+1)^2$

$=$

$12(x-1)$

Since the coefficient of

$(x-1)$ is positive, use

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

This also means that the parabola will be concave right

Find the focal length ( $a$ ) and vertex by comparing the given equation to the standard form

${(y-\color{#007ddc}{k})}^2$	$=$	$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$
${(y-\color{#007ddc}{(-1)})}^2$	$=$	$12(x-\color{#00880a}{1})$
$(y+1)^2$	$=$	$12(x-1)$

The Vertex

$($ $h$

$,$ $k$

$)$ can be read from the equation

$($ $1$

$,$ $-1$

$)$

Focal Length:

$4a$	$=$	$12$
$4a$ $divide4$	$=$	$12$ $divide4$	Divide both sides by $4$

$a$	$=$	$3$

Identify the focus and directrix.

Focus:

$($ $a$

$+$ $h$

$,$ $k$

$)$ becomes

$($ $3$

$+$ $1$

$,$ $(-1)$

$)$ or

$(4,-1)$

Directrix:

$x=-$ $a$

$+$ $h$ becomes

$x=-$ $3$

$+$ $1$ or

$x=-2$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave right

Question 4 of 4

Sketch the graph of

$x=1/4(y^2-6y+13)$

1.
2.
3.
4.

Incorrect

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Standard Form (Concave Right)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$($ $a$

$+$ $h$

$,$ $k$

$)$

Standard Form (Concave Left)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$(-$ $a$

$+$ $h$

$,$ $k$

$)$

First, transform the given equation into vertex form.

$x$	$=$	$1/4(y^2-6y+13)$

$x$ $times4$	$=$	$1/4(y^2-6y+13)$ $times4$	Multiply both sides by $4$

$4x$	$=$	$y^2-6y+13$
$y^2-6y$	$=$	$4x-13$	Keep only $y$ terms on the left side
$y^2-6y$ $+9$	$=$	$4x-13$ $+9$	Complete the square by adding to both sides: $(6divide2)^2=9$
$(y-3)^2$	$=$	$4x-4$	Contract left side
$(y-3)^2$	$=$	$4(x-1)$	Factor right side

First, determine which standard form applies to the given equation.

$(y-3)^2$

$=$

$4(x-1)$

Since the coefficient of

$(x-1)$ is positive, use

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

This also means that the parabola will be concave right

Find the focal length ( $a$ ) and vertex by comparing the given equation to the standard form

${(y-\color{#007ddc}{k})}^2$	$=$	$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$
${(y-\color{#007ddc}{3})}^2$	$=$	$4(x-\color{#00880a}{1})$

The Vertex

$($ $h$

$,$ $k$

$)$ can be read from the equation

$($ $1$

$,$ $3$

$)$

Focal Length:

$4a$	$=$	$4$
$4a$ $divide4$	$=$	$4$ $divide4$	Divide both sides by $4$

$a$	$=$	$1$

Identify the focus and directrix.

Focus:

$($ $a$

$+$ $h$

$,$ $k$

$)$ becomes

$($ $1$

$+$ $1$

$,$ $3$

$)$ or

$(2,3)$

Directrix:

$x=-$ $a$

$+$ $h$ becomes

$x=-$ $1$

$+$ $1$ or

$x=0$

Start graphing the parabola by plotting the vertex, focus and directrix

Finally, draw a parabola concave right

Graphing Parabolas Given in General Form

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Quiz summary

Information

1. Question

Hint

Excellent!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

2. Question

Hint

Well Done!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

3. Question

Hint

Nice Job!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

4. Question

Hint

Fantastic!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

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